When a mixture of N2 and H2 in the vol ratio of 1:5 is allowed to react at 100K and 10^3 atm pressure, 0.426 mole fraction of NH3 is formed at equilibrium.Find Kp.

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To find the equilibrium constant \( K_p \) for the reaction between nitrogen (N2) and hydrogen (H2) to form ammonia (NH3), we can follow a systematic approach. The balanced chemical equation for the reaction is:

N2 + 3H2 ⇌ 2NH3

Given the volume ratio of N2 to H2 as 1:5, we can start by determining the initial moles of each gas. Let's assume we have 1 mole of N2; therefore, we have 5 moles of H2. This gives us a total of 6 moles in the mixture.

Initial Moles and Mole Fractions

From the initial conditions, we can summarize the moles as follows:

• N2: 1 mole
• H2: 5 moles
• Total: 6 moles

The initial mole fractions are:

• X(N2) = 1/6
• X(H2) = 5/6

Change in Moles at Equilibrium

At equilibrium, we know that the mole fraction of NH3 is 0.426. This means that the total moles at equilibrium can be expressed in terms of the moles of NH3 formed. Let \( x \) be the number of moles of NH3 formed. According to the stoichiometry of the reaction, for every 1 mole of N2 consumed, 3 moles of H2 are consumed, and 2 moles of NH3 are produced. Thus, we can express the changes in moles as follows:

• N2: 1 - \( \frac{x}{2} \)
• H2: 5 - \( \frac{3x}{2} \)
• NH3: \( x \)

Since the mole fraction of NH3 is given as 0.426, we can express this in terms of the total moles at equilibrium:

Calculating Total Moles at Equilibrium

The total moles at equilibrium can be expressed as:

Total moles = \( (1 - \frac{x}{2}) + (5 - \frac{3x}{2}) + x = 6 - x \)

Now, we can set up the equation for the mole fraction of NH3:

0.426 = \( \frac{x}{6 - x} \)

Solving for x

Cross-multiplying gives:

0.426(6 - x) = x

2.556 - 0.426x = x

Combining like terms results in:

2.556 = 1.426x

Thus, solving for \( x \):

x = \( \frac{2.556}{1.426} \approx 1.794 \) moles of NH3 formed.

Calculating Moles of Reactants at Equilibrium

Now we can find the moles of N2 and H2 at equilibrium:

• N2: \( 1 - \frac{1.794}{2} \approx 0.103 \) moles
• H2: \( 5 - \frac{3 \times 1.794}{2} \approx 0.611 \) moles

Finding Kp

The equilibrium constant \( K_p \) is defined as:

\( K_p = \frac{(P_{NH3})^2}{(P_{N2})(P_{H2})^3} \)

To find the partial pressures, we can use the ideal gas law, where \( P = \frac{nRT}{V} \). However, since we are given the total pressure (10^3 atm), we can directly relate the mole fractions to the partial pressures:

• P(N2) = \( \frac{0.103}{6} \times 10^3 \approx 17.17 \) atm
• P(H2) = \( \frac{0.611}{6} \times 10^3 \approx 101.83 \) atm
• P(NH3) = \( \frac{1.794}{6} \times 10^3 \approx 299.00 \) atm

Now substituting these values into the \( K_p \) expression:

\( K_p = \frac{(299.00)^2}{(17.17)(101.83)^3} \)

Calculating this gives:

\( K_p \approx 0.045 \) (after performing the calculations).

Thus, the equilibrium constant \( K_p \) for the reaction at the given conditions is approximately 0.045. This value indicates the extent to which the reaction favors the formation of products at equilibrium under the specified conditions.